Finding the type of a pointer

Question

I want to do something like this:

SomeType *y;

/* ... snip ... */

auto x = new decltype(y); // Create a new pointer "x" to a SomeType object.

But decltype(y) is SomeType* and decltype(*y) is SomeType&. Is there a way to get plain SomeType out of y?

Answer 1

Since decltype(*y) is a reference, you can use std::remove_reference:

new std::remove_reference<decltype(*y)>::type;